2006-06-30 09:42:52 · 21 answers · asked by Useless 5 in Entertainment & Music ➔ Horoscopes
It is not painful for the person whose birthday. But as the mother of a child born on the cusp of Taurus/Gemini. IT IS PAINFUL! She is moodiest of my three.
2006-06-30 09:48:05 · answer #1 · answered by Bird 2 · 2⤊ 3⤋
No, being born on a cusp means that you are born between two astrological signs. For example, if you were born March 20th you would be on the cusp of Aries and Pisces. Some say that Aries starts March 21 while you might find that in others it starts March 20.
2006-06-30 16:45:51 · answer #2 · answered by ♥Cute City Chick♥ 4 · 0⤊ 0⤋
no i think it a good thing i was born on march 20th and i have traits of aris and piscean but cuz im piscean and its at the end of the zodica then i have traits from all star signs so count ur self lucky u were only born on a cusp and not on a cusp and on the end of a zodica lmao
2006-07-05 01:58:57 · answer #3 · answered by donna 1 · 0⤊ 0⤋
Being born at any time is painful both for baby and mother!!!!! Apparently being born is the most stressful thing we will ever experience in our entire lives
2006-06-30 18:12:17 · answer #4 · answered by Eleanora 3 · 0⤊ 0⤋
no it is not pain full to be born on a cusp.
2006-06-30 16:46:59 · answer #5 · answered by flori 4 · 0⤊ 0⤋
Why would it be painful... I'm born on the 21st June and I fortunately have traits of both a Gemini and a Cancerian
2006-06-30 16:46:47 · answer #6 · answered by isseybarbie 3 · 0⤊ 0⤋
No cos it's just 2 starsigns, which mean NOTHING, unless you want 2 put your faith and trust in random starsigns which mean very little, if not nothing at all!
2006-07-01 19:18:01 · answer #7 · answered by CozOfGrace 3 · 0⤊ 0⤋
Yes - OUChhhhhhhhhhh
2006-07-03 06:00:48 · answer #8 · answered by Suzita 6 · 0⤊ 0⤋
what do you think about this date : 29 Feb 1980
2006-07-02 10:27:34 · answer #9 · answered by Anonymous · 0⤊ 0⤋
It's a bit like being picked up by the fuzz.( police ) Dont arf. make yours eyes water.
2006-06-30 17:23:11 · answer #10 · answered by robert c 3 · 0⤊ 0⤋